Set f(x) = x³ - 2x² + 10x + 136.
Use Descartes' rule of signs.
f(x) has 2 change in sign, so there are
(1) 2 real roots, or
(ii) a pair of conjugate imaginary roots.
f(-x) = -x³ - 2x² - 10x + 136
f(-x) has 1 sign change so there is 1 real root.
Scan the function for the first zero crossing, using negative values of x.
x f(x)
--- ---------
-1 123
-2 100
-3 61
-4 0
x = -4 is a root, so (x+4) is a factor.
Perform synthetic division (or long division if you do not know synthetic division).
-4 | 1 -2 10 136
-4 24 -136
---------------------
1 -6 34 0
Therefore
f(x) = (x+4)(x²- 6x + 34) = 0
Before factoring x² - 6x + 34, test the discriminant.
D = (-6)² - 4*34 = -100 => we have a pair of imaginary roots.
Use the quadratic formula.
x = (1/2)[6 +/- √-100]
= 3 +/- 5i
Answer: C
The roots are -4, 3 +/- 5i
The answer would be 1,600 because it’s a fraction
Answer: 8 + 49 factored is 57
I hope this helps you
-3
Because -2^2-8+1
-2to the second power is 4
4-8=-4
-4+1=-3
Answer: 40
Step-by-step explanation:
10 + 10 + 10 + 10 = 40
